[Math] $100$ consecutive natural numbers with no primes

elementary-number-theorysequences-and-series

I have already seen a duplicate of this. But I was not able to follow along.

So I was asked this question by my maths teacher during a sequence and series lecture.

  1. Can you devise $100$ consecutive natural numbers with no primes.

  2. Additionally, Is it possible to have $1000$ consecutive natural numbers with exactly $12$ primes between them?

I have an intuition that we have to form a recurrsive relation and solve it. But I am stuck.

I also tried making a $10*10$, having $50$ multiplies of $2$, $33$ multiples of $3$ and so on trying to generalize but wasn't able to come up with a solution.

Any hints would be really helpful. Thank you.

Best Answer

  1. $101!+2,101!+3,...,101!+101$ has no primes (because $i|101!+i$ for $2\leq i\leq 101$)
  2. Let $f(n)$ be the number of primes in $\{n,...,n+999\}$. Since $f(1)>12$ and $f(1001!+2)=0$ (for the same reason as above) and $|f(i+1)-f(i)|\leq1$, there exists $i\in (1,1001!+2)$ with $f(i)=12$.
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